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I can’t get to my maths Tutor due to this CORONNAAAA
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Answers
Given : A three Digit number is 75 times the average of sum of Digit
To Find : Possible numbers & possible sum of Digits of Such numbers
Solution:
Let say 3 Digit number = ABC
Then Value of number = 100A + 10B + C
Sum of Digits = A + B + C
Average of Digits = (A + B + C)/3
Number is 75 times the Sum of Digit
=> 100A + 10B + C = 75 (A + B + C)/3
=> 100A + 10B + C = 25 (A + B + C)
Now any number of multiple of a number ending with 5 can end with 5 or 0
Case 1 : C = 0
=> 100A + 10B = 25(A + B)
=> 75A = 15B
=> 5A = B
A = 1 , B = 5 ( A = 2 will lead B = 10 not possible)
150 is the number
Average of Digits = ( 1 + 5 + 0)/3 = 2
2 * 75 = 150
Case 2 : C = 5
=> 100A + 10B + 5 = 25 (A + B + 5)
=> 75A = 15B + 120
=> 5A = B + 8
A = 2 , B = 2
A = 3 , B = 7
A = 4 , B = 12 ( not possible)
225 Average = (2 + 2 + 5)/3 = 3 & 3 *75 = 225
375 Average = (3 + 7 + 5)/3 = 5 & 5 *75 = 375
150 , 225 & 375 are Numbers
Possible Values of Sum Of Digits
1 + 5 + 0 = 6
2 + 2 +5 = 9
3 + 7 + 5 = 15
150 , 225 & 375 are Numbers
6 , 9 & 15 are possible Sum of Digits of number
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